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Transcript of L7 ReactorTheory - Citadelece.citadel.edu/barsanti/elec307/L7_ReactorTheory.pdfbare reactor as for...
4/14/2015
1
Nuclear Reactor Theory
Chapter 6
1
Reactor Theory1. One Group Reactor Equation
a) Assumptions bare reactor
(1) Rx is a homogeneous fuel & moderator
(2) Rx consists of one region and has neither a
blanket or reflector.
b) Diffusion Equation
∅ − Σ∅ + =
Eq. 1
D – Diffusion coefficient
Σ - x section for absorption
v- n yield
S- source density (n/cm3 –s) = νΣϕ2
Reactor Theory
If the fission source neutrons do not balance the
leakage and absorption terms, then the RHS of Eq. 1 is
non-zero.
In this case we could balance the equation by
multiplying the source term by a constant say “1/k ”.
If the source is too small then k <1.
If the source is too large then k >1.
3
Reactor Theory
So we have
∅ − Σ∅ +1Σ∅ = 0
Let =
Σ - Σ
then ∅ = −∅
substituting −∅ for ∅
We get −∅ − Σ∅ +
Σ∅ = 0 Eq.2
Solving for k : =∅
∅ !∅=
!
4
Reactor Theory
k = the multiplication factor
=∅
∅ !∅=
!
Numerator = neutrons born via fission
Denominator = neutrons lost to leakage and
absorption.
5
Reactor Theory
Notice that = νΣϕ = ηΣ#ϕ = η!$!Σϕ
#n/fiss Fuel f
n per abs
f = !$!
= fuel utilization: the fraction of all
neutrons absorbed in the reactor that are
absorbed in the fuel.
= η%Σϕ Eq. 3.6
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2
Reactor Theory
C. Multiplication factor in the infinite reactor.
Assumptions:
1. All neutrons are absorbed. No neutrons leak.
2. The neutron flux φ is constant everywhere. Is
independent of position.
Concept:
1. Since all neutrons born are eventually
absorbed, then Σϕ is the total number of
neutrons.
2. Of these %Σϕ are absorb in the fuel, and
release η%Σϕ in the next generation 7
Reactor TheoryMultiplication factor in the infinite reactor.
Concept:
3. Dividing the number of neutrons in one generation
by those in the next gives effective yield/absorp
& ='!(
!(= η%
4. & is k for and infinite reactor. Fuel utilization
5. Since η and % are constants that depend on the
material properties of the reactor, & is the same for a
bare reactor as for an infinite reactor of the same
composition.
6. Thus & refers to a Rx in which no neutrons leak.
8
Reactor TheoryD. Buckling B2 in the Critical bare reactor.
1. Since & ='!(
!(= η%
2. And the source term = η%Σϕ,then =&Σϕ
3. Using this in the one-group Rx eqn. use Eqn. 2 & 3
−∅ − Σ∅ +1 =
*+*,
−∅ − Σ∅ +1(&Σ∅) =
*+*,
4. If Rx is critical, k = 1, and
= 0,
−∅ + (&−1)Σ∅ =09
Reactor TheoryD. Buckling B2 in the Critical bare reactor.
4. If Rx is critical, k = 1, and
= 0,
−∅ + (&−1)Σ∅ =0
5. Let / = !0 = diffusion area, then
−∅ +(& − 1)∅
/= 0
6. Solving for B2
=(& − 1)/
10
Reactor Theory
Ex 6.1 … Find f and k∞for a mixture of U-235 and sodium
(Na) in which the U is present to 1 w/0.
Solution: % =!$!=
!$!$ !2
=
3!2 3!$0
=
425!24$5!$
67&6#=atomic concentrations of Na and U in 9:;<=/?9@
So A2A$=
B2C2
C$B$
where DE=F<+=E,GH
IJ@
and LE=MNO9O,:9E?PM,H JQRS⁄
11
Reactor Theory
Now: %U = 1 +B2B$
C$C2
V!2V!$
But : B$
B$ B2= 0.01 ⇒
B2B$= 99
Thus: %U = 1 + 99@Z
@
[.[[[\
.]Z= 1.48
So: % = 0.671
Next: & = +% = 2.2 0.671 = 1.48
Thus an infinite reactor of this composition would
be super critical.12
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3
Reactor Theory
II. Solutions to the one group reactor equation.
The Slab Reactor
Assume:
1. Infinite bare reactor of thickness ‘a’.
2. Is critical (flux in steady-state)
-a/2 0 a/2 x
a
13
Reactor Theory
2. Rx Eqn. is c∅
cd+ ∅ = 0
With the general solution ∅ e = f cos e + f sin eWhere C1 and C2 are to be determined using Ficks law.
3. l = −c∅
cd= 0 at x = 0 due to symmetry of slab, so
c∅
cd= 0.
Thus: c∅
cd= −f sin 0 + f cos 0 = 0O,e = 0
So: f = 0! and: ∅ e =ff:=(e)
Since: ∅
= f cos
= 0 and =
n
+ = 1,3,5,…
Finally: ∅ e =f cos n
e + = 1,3,5,…
14
Reactor Theory
The Slab Reactor
∅ e =f cos+rOe + = 1,3,5,…
For a critical reactor only the n = 1 eigenvalue will exist.
∅ e = f cos+rOe
-a/2 0 a/2 x
n = 1
n = 3
15
Reactor Theory
The Slab Reactor
The value of C1 depends on the reactor power level (P).
From the text page 274,
f =n
st!u
P = power per unit area v7 IJw⁄
xy =constant conversion factor of 3.2 x 10U
∅ e = n
st!u?:=(
n
)
16
Reactor Theory
B. Other Reactor Shapes
1. Sphere of radius R
∅(N) =u
4xyΣzsin(rN z0 )
z
2. Infinite Cylinder of radius R
∅ N =[.@\|
stylQ(
.[Z~
y)
3. Solutions for other shapes listed in table 6.2.
17
Reactor Theory
C. Buckling
Recall the solution for the slab reactor
∅ e = fcos( e)The square of the first eigenvalue is called the
buckling of the reactor.
Since c ∅
cd+
∅ = 0
Then =
U
∅
c ∅
cd
The RHS is an expression that is proportional to the
curvature of the flux, which in turn is a measure of
how much the flux curves or “buckles”. 18
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4
Reactor Theory
C. Buckling
Notice that = (
n
) for a slab, and that
decreases as ‘a’ increases. In the limit as ⇒ ∞,
=0, φ = constant and thus has no buckle.
For the reasons listed below, it is important that the flux distribution in an actual reactor be as flat as possible..
– Reduced power spikes
– Uniform fuel burn-up
– Prevent ‘hot channel’ effects 19
Reactor Theory
D. Max. to Avg. Flux Ratio Ω (table 6.2)
Ω =∅Jd∅H
is a measure of the variation of the flux w/ in a reactor,
and the extent to which the max. power density at the
center exceeds the avg power density.
e.g. For a bare infinite slab: ∅ e = f cos n
e ,
so ∅Jd= f
And ∅H = fcos( nd
/U/ )Fe =
n
Thus Ω =
0=n
= 1.57
20
Reactor Theory
E. Ex 6.2. A bare spherical rx radius R = 50 cm, operates at 100Mw=10\J/s and Σ =0.0047?9U.
What are the max and avg. flux values?
∅Jd = lim~→[|
sty(
t)
~=
|
styn
y
= [n
@.d[ [.[[(Z[)w= 4.18e10Z
IJU7
∅H =.\d[
@.= 1.27e10Z
IJU7Table 6.2
21
Reactor Theory
III. k for a non- infinite reactor
Recall a critical reactor has:
=U
which was found for &=1
rearranging :
= 1 Eqn. 1
Since n’s either leak or are absorbed, the
relative probability that a n will be absorbed is
called the non-leakage probability u .22
Reactor Theory
III. k for a non- infinite reactor
u =#:%+=O=:N<F
#:%+=O=:N<F:N;<O<F=
Σ∅
Σ∅ + ∅
=!
! 3!0
3!0=
u =
Eqn. 2
Comparing Eqn. 1 and 2
&u = 1%:NO?NE,E?O;N<O?,:N.
23
Reactor Theory
III. k for a non- infinite reactor
For other then critical conditions we can write:
= &u
24
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5
Reactor Theory
III. k for a non- infinite reactor
Ex…Given a reactor bare sphere of radius R = 48.5
cm, with / = 384?9.
What is the probability that a fission neutron
will be absorbed?
u =1
1 + /=
1
1 + 384(r 48.5)0 = 0.38
table 6.2
38% chance of absorption, 62% chance of leakage.
25
Reactor Theory
IV. Thermal Reactors
A. Recall that thermal rx’s contain a moderator to slow down fission neutrons to thermal energies. For convenience all materials in the reactor other than fuel are considered moderator.
B. Thermal Utilization (f) formerly the fuel utilization is the fraction of all neutrons absorbed that are absorbed in the fuel in a thermal reactor.
% =ΣΣ
26
Reactor Theory
IV. Thermal Reactors
C. Fast fission factor (ε) is defined as the ratio of
the total number of fission neutrons produced
by both fast and thermal fission to the number
produced by thermal fission alone. The value of
ε (fast fission factor) ~ 1.02- 1.08. Thus about 2 –
8 % of fission are from fast neutron reactions.
27
Reactor Theory
IV. Thermal Reactors
D. The resonance escape probability (p) is the probability that a neutron is not absorbed while slowing down by nuclides having absorption resonances above thermal energies.
E. Based on previous discussion, the absorption of Σ∅thermal neutrons leads to the production of η%Σ∅ new neutrons in an infinite thermal reactor.
The infinite rx multiplication factor is;
& =η%Σ∅Σ∅
= η%ε
28
Reactor Theory
IV. Thermal Reactors
F. Although in a thermal reactor most fissions
occur with neutrons at thermal energies, all
fission neutrons are born at fast energies.
Therefore it is customary to describe a thermal
reactor by two groups of neutrons (fast and
thermal). Thus there are to groups of neutron
flux terms:
Let ∅ = ∅,<%O=,+<,N:+%;ex+1
And
∅ = ∅,<,<N9O;+<,N:+%;ex+229
Reactor Theory
IV. Thermal Reactors
Substituting and rearranging the diffusion
equation leads to (see page 289 of text)
The Two Group Critical Equation
(
)( )= 1
30
P<N</ =
Σ= ,<N9O;FE%%=E:+ON<O
O+F =Σ= +<,N:+OM<
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6
Reactor TheoryIV. Thermal Reactors
/<,,E+Mu =
( )
= thermal non-leakage
probability
O+Fu# =
( )= fast non-leakage probability
Which leads to &uu# = 1for a critical reactor.
Or = &uu# = η%εuu#Which is the six factor formula for a thermal reactor!
31
Reactor Theory
IV. Thermal Reactors
Neutron life cycle = η%εuu#
32
Σ#∅ → η → ¡ → u# → → u → % →
n’s via
thermal
fission
n’s from
thermal + fast
fission
Fast n’s
that don’t
leak
Slow n’s
Slow n’s
that don’t
leak Slow n’s
absorb in
fuel
n’s
absorbed
in fuel
33 34
35 36
4/14/2015
7
Reactor Theory
IV. Thermal Reactors
37
Reactor Theory
IV. Thermal Reactors
38
Reactor Theory
IV. Thermal Reactors
39
Reactor Theory
IV.
40
Reactor Theory
IV.
41
Reactor Theory
IV.
42